How can I get Cartesian products of some subsets of my lists, as well as the product of the entire list of lists?

Question

It's probably on Google, but I'm having a hard time putting into words what I'm trying to do. itertools.product is close, but I also want it to give me smaller combinations whereas itertools only gives combinations across all lists. It would probably be easier to explain in an example:

Example of lists I want to combine: [[a,b,c],[d,e,f],[g,h,i]]

Itertools gives back: [a,d,g],[a,d,h]....[c,f,h],[c,f,i] (27 results)

Output I am looking for: The above, plus:

[a,d],[a,e],[a,f],[b,d],[b,e],[b,f],[c,d],[c,e],[c,f],[d,g],[d,h],[d,i],[e,g],[e,h],[e,i],[f,g],[f,h],[f,i]

There would be a total of 45 results. Note that I'm only looking for combinations between the lists that are next to each other (don't want [a,g] returned). The lists will not be this simple either, there will be a set of 10 lists with 3-4 characters each.

The only line of code I have for this is: list(itertools.product(*thelist))]. I just don't know where to go next on what I'm trying to do here. I have a feeling there's some way to do this with itertools, I just can't figure it out.

I tried briefly, but was having difficulty, maybe wasn't implementing it right with the 2D lists. I'm going to play with itertools.combinations some more. — SCAND1UM
– SCAND1UM, Commented Mar 24, 2021 at 21:26

Luke Woodward · Accepted Answer · 2021-03-24 21:43:46Z

It looks to me like you want to run itertools.product on each sublist of your list of lists, and concatenate the results of that together.

Does the following do what you need?

import itertools

def combine(lists, min_length):
    items = []
    for start in range(len(lists)):
        for end in range(start + min_length, len(lists) + 1):
            items.extend(itertools.product(*lists[start:end]))
            
    return items

I called this function using the following:

combine([["a","b","c"],["d","e","f"],["g","h","i"]], 2)

and it returned a list of 45 tuples, which appeared to contain all of the items you are looking for.

pakpe · Accepted Answer · 2021-03-24 21:48:06Z

You have the first 27. Here is how you get the other 18:

import itertools as it
lst = [['a','b','c'],['d','e','f'],['g','h','i']]

result = [it.product(lst[i], lst[i+1]) for i in range(len(lst)-1)]
result = list(it.chain.from_iterable(result))

print(result)
#[('a', 'd'), ('a', 'e'), ('a', 'f'), ('b', 'd'), ('b', 'e'), ('b', 'f'), ('c', 'd'), ('c', 'e'), ('c', 'f'), ('d', 'g'), ('d', 'h'), ('d', 'i'), ('e', 'g'), ('e', 'h'), ('e', 'i'), ('f', 'g'), ('f', 'h'), ('f', 'i')]

Iluvatar · Accepted Answer · 2021-03-24 22:19:31Z

Using pretty clever bit checking method from HERE

from itertools import product

x = ('abc', 'def', 'ghi')

def subsets(s):
    sets = []
    for i in range(1, 1 << len(s)):
        if bin(i).count("1") != 1 and '11' in bin(i):
            subset = [s[bit] for bit in range(len(s)) if is_bit_set(i, bit)]
            sets.append(subset)
    return sets

def is_bit_set(num, bit):
    return num & (1 << bit) > 0

a = 0
super_sets = subsets(x)
for sset in super_sets:
    a += len(list(product(*sset)))
    #print(list(product(*sset)))
print(a)

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